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Suppose I have a morphism f:X→Y such that the relative sheaf of differentials ΩX/Y is locally free. Does it follow that f is smooth?

The answer is no, but for a silly reason. You could have some non-reducedness (Spec(k[e]/(e^2)) over Spec(k) has a free sheaf of differentials, but isn't smooth). But what if you add the hypothesis that the rank of ΩX/Y is dim(X)-dim(Y)?

Edit: As Jonathan points out in his answer, I was careless with my counterexample. It only works if char(k)=2.

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3 Answers 3

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I think Ishai's example is close, but one must be a little careful; the normalization of the node is a good example, but the normalization of the cusp is ramified, and the sheaf of relative differentials in that case is not even locally free.

The differential-wise condition you want is this: for the morphism morphism f: X --> Y to be smooth, you need that the sequence

0 --> f^* Omega_Y --> Omega_X --> Omega_X/Y --> 0

be exact and locally split (I can't find a reference that says this is sufficient, so it may not be). In the special case when dim X = dim Y, Omega_X/Y is 0 if and only if f is unramified. But in this case f^* Omega_Y --> Omega_X can still fail to be injective.

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How do you see that the normalization of the cusp is ramified? What's wrong with Ishai's calculation that shows that the relative sheaf of differentials of A~ over A is zero? –  Anton Geraschenko Oct 1 '09 at 16:57
    
Fiber the normalization with the cusp. You get something non-reduced. (You can see this with explicit equations.) –  David Zureick-Brown Oct 2 '09 at 10:15
    
Ok, now I understand what's wrong with Ishai's calculation. I'll post a comment there. –  Anton Geraschenko Oct 3 '09 at 1:49

Let X = spec A be an affine integral scheme of dimension one which is not smooth. Smoothness may be checked smooth-locally on the source (given U --> V if there exists a W --> U which is smooth and surjective such that W --> V is smooth then U --> V was smooth). Thus, the normalization X~ = spec A~ cannot be smooth over X. But if d: A~ --> M is any derivation of A~ over A and a/b is an element of A~ then d(a/b)= (bda - adb)/(b^2)= 0; this shows that the rel. differentials are zero, hence in particular loc free of fin rank.

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Ok, it turns out there's a problem with the calculation d(a/b)=(bda - adb)/(b^2)=0. That works for showing that the relative differentials of the fraction field over the ring are zero, but the coefficients of the right hand side of that calculation may not be in the normalization. As Charley pointed out to me, the normalization of the cusp, Spec(k[x,y]/(y^2-x^3)) is Spec(k[y/x]), but you can't write d(y/x)=(xdy-ydx)/x^2 because x/x^2 and y/x^2 are not in the normalization. –  Anton Geraschenko Oct 3 '09 at 1:53

A variation on Ishai's example is a closed embedding: its sheaf of relative differentials is 0, hence free of finite rank, even though it needn't be smooth.

However, k[e] / e^2 over k is not actually a counterexample (except in characteristic 2). The module of relative differentials of Spec k[e] / e^2 over Spec k is not free if the characteristic of k is not 2. Let A = k[e] and B = k[e] / e^2. Then

Omega_B = Omega_A (x) B / d(e^2) = k[e] / (e^2, 2e)

via the isomorphism Omega_A --> A : dt --> 1. This is not isomorphic to B unless 2 = 0.

On the other hand, you can conclude that B is smooth if its cotangent complex is a vector bundle in degree 0. In the case of k[e] / e^2, the cotangent complex is

[ I_{B/A} / I_{B/A}^2 ---> Omega_A (x) B ] = [ e^2 A / e^4 A ---> B de ]

in degrees [-1,0] and the differential is the universal derivation. (I write I_{B/A} for the ideal of B in A.) Even in characteristic 2, the differential has a kernel, so the cotangent complex is not concentrated in degree 0.

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Thanks for the correction. I think that a closed immersion gives the same sort of counterexample as k[e]/(e^2) in characteristic 2. The relative sheaf of differentials is free, but of rank 0, rather than dim(X)-dim(Y), which is negative. –  Anton Geraschenko Oct 1 '09 at 14:35
    
A closed immersion doesn't have to have negative relative dimension. –  Jonathan Wise Oct 1 '09 at 16:49
    
@Jonathan: Oh, right, you could have the inclusion of an irreducible component. Good point. –  Anton Geraschenko Oct 1 '09 at 21:14

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